Download Design Guidelines for Bipolar Transistor Audio Preamplifier Circuits

Design Guidelines for Bipolar Transistor Audio Preamplifier Circuits
By Mike Martell
N1HFX
Before attempting to design a transistor
amplifier circuit, it is necessary to acquaint
ourselves with some very important design
equations. The most commonly used design
equations are listed to the right to help us with
our effort. The first few equations are derived
from ohms law and you should already be
familiar with them. The bottom two equations
deal with transistor gain and are equally
important to our work. The formula for hfe
refers to the ratio of collector current (Ic) to
base current (Ib). For most modern transistors
it is typically in the 50 to 100 range. To insure
a circuit will always work properly, it is safe to
assume a value of 50 for modern transistors.
At this time, we need to make some decisions
about our audio preamplifier circuit. For this
design example we will choose the following:
•
•
•
•
Vcc = 12 V
Ic = 5 ma
hfe = 50
Q1 = 2N3904
With all those big decisions made, we can
now begin our design.
Using the one of the hfe formulas, we will now calculate the base current as follows:
Ib = Ic / hfe
Ib = .005 / 50 = .1 ma
Lets calculate R1, the collector load resistor, as follows:
R1 = 1/2Vcc / .005
R1 = 6 / .005 = 1200
Notice above that we assumed 1/2 of the supply voltage to be dropped across R1. This is
necessary to insure that the amplifier remains in the linear operating range of the transistor.
We need to determine how much voltage is to appear across the emitter resistor, R2, before we
calculate its value. A good value is anywhere between 5 to 10 percent of Vcc. For this circuit we
will use 1 volt which is about 8 percent of Vcc. Resistor R2 is now calculated as follows:
R2 = 1 / (Ic + Ib)
R2 = 1 / (.005 + .0001) = 196
Because the voltage across the base to emitter of a silicon transistor is always .7 volts, the
voltage from the base to ground is .7 plus the 1 volt drop across R2 for a total of 1.7 volts. This
1.7 volts happens to be the voltage drop across resistor R4. In order to provide a stiff base
voltage, resistor R4 should have a current of about 5 to 10 times the base current. For this
example, we will assume 9 times the base current for a total of .9 ma. Resistor R4 can now be
calculated as follows:
R4 = 1.7 volts / .0009 = 1889
If the voltage drop across R4 is 1.7 volts then the voltage drop across R3 must be 12 - 1.7 for a
total of 10.3 volts. The current through R3 is the total of the current through R4 (.9ma) and the
base current (.1ma) for a total of 1 ma. R3 can now be calculated as follows:
R3 = 10.3 / .001 = 10300
Now that we have calculated all our resistor values, we will select the nearest standard values as
indicated below:
R1 = 1.2K
R2 = 180
R3 = 10K
R4 = 1.8K
The circuit at right is the result of our design
efforts. 4.7uF capacitors were use for input
and output coupling and slightly larger or
small values could be used satisfactorily.
Notice the optional 4.7uf capacitor across the
emitter resistor R2. This capacitor increases
the current gain to the hfe of the particular
transistor used. This emitter bypass capacitor
should only be used when the maximum
amount of gain is desired without regard to a
predictable level of gain.
Remember, hfe will vary from transistor to transistor even though they have the same part
number and even if they were produced by the same manufacturer. It is always better to assume
an hfe that is at least 20% less than that specified by the manufacturer.
One common error that designers make is that they forget to calculate the actual power that each
resistor will dissipate in a circuit. Failure to perform these calculations can sometimes result in a
resistor exceeding its maximum power level and cause premature resistor failure. This is
particularly important for circuits which have collector currents exceeding 40 milliamps.
Fortunately, power ratings for each of our resistors in this circuit can be easily calculated as
follows:
R1 = 6 Volts * .005 Amps = .03 Watts
R2 = 1 Volt * .0051 Amps = .0051 Watts
R3 = 10.3 Volts * .001 Amps = .0103 Watts
R4 = 1.7 Volts * .0009 Amps = .00153 Watts
The preceding calculations indicate that 1/4 watt, 5% resistors are adequate for this design.
Although our preceding circuit does have substantial gain, lets design a second stage to the
previous circuit to further increase gain. Before we can begin our design we must make those all
important design decisions again as indicated below:
Q1 = 2N3904
Vcc = 12V
Ic = 10ma
hfe = 50
Vr2 = 1V (8% of Vcc)
Ir4 = 9 times Ib
Now were ready to calculate our resistor values as follows:
R1 = 6/.010 = 600
Ib = .01/50 = .0002
R2 = 1/(Ic + Ib) = 1/(.01 + .0002) = 98
Ir4 = 9 * .0002 = .0018
R4 = (1 + .7)/.0018 = 944
R3 = (12 - 1 - .7)/(.0018 + .0002) = 10.3 / .002 = 5150
We will select standard resistor values as
follows:
R1 = 560
R2 = 100
R3 = 5.1K
R4 = 1K
The circuit at right is the result of our design
efforts. 10uF capacitors were use for output
coupling and optional emitter bypass. Slightly
larger or small capacitor values could also be
used satisfactorily. As always, the emitter
bypass capacitor should only be used when
the maximum amount of gain is desired
without regard to a predictable level of gain.
Now lets add the two transistor stages together to get the resulting circuit below:.
Because of the large amount of gain obtained with this circuit, we added a 10K variable resistor at the
output as a gain control. This circuit will provide good results for almost any microphone pre-amplifier
application. Incidentally, we can substitute a 2N2222A, 2N4401 or any general purpose NPN with a
minimum hfe of 50 for the 2N3904 transistor used in this design example. There are many general
purpose NPN transistors that can be easily used in this circuit with good results.
Perhaps a good exercise for the designer is to recalculate all of the above resistor values using a Vcc
of 9 volts which will allow the use of a 9 volt battery.
Now that we got our hands wet designing these circuits, lets summarize what we've learned below:
Vcc
hfe
Ic
Ib
= Supply Voltage
= Absolute minimum current gain for the selected transistor
= Selected collector current
= Ic / hfe (base current)
Vr1
Vr2
Vr3
Vr4
= 1/2 Vcc
= 5 to 10% of Vcc
= Vcc - .7 - Vr2
= .7 + Vr2
Ir4
= 5 to 10 times Ib
R1
R2
R3
R4
= 1/2Vcc / Ic
= Vr2 / (Ic + Ib)
= (Vcc - .7 - Vr2) / (Ir4 + Ib)
= (.7 + Vr2) / Ir4
DE N1HFX